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December 18, 2014

December 18, 2014

Posted by **lynn** on Tuesday, April 30, 2013 at 11:56am.

- Algebra -
**Reiny**, Tuesday, April 30, 2013 at 1:12pmsimplified ...

f(x) = (x^2 + 1)(x-1)(x+1)/(3x(x-1))

= (x+1)(x^2+1)/(3x)

= (x^3 + x^2 + x + 1)/(3x)

= (1/3) (x^2 + x + 1) + 1/(3x)

clearly there is a vertical asymptote at x = 0

there is also a "hole" at (1, 4/3) , (we divided top and bottom by x-1 )

when x --> ∞

1/3x --> 0

so we left with

(1/3)(x^2 + x + 1) + 0 , which is a paraloba

so we have a "curved" asymptote formed by the parabola

y = (1/3)(x^2 + x + 1)

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